Question

A married couple agreed to continue bearing a new child until they get two boys , but not more than 4 children. Assuming that each time that a child is born, the probability that is a boy is 0.5 independent from all other times. Find the probability that the couple has atleast two girls. A. 1/2 B. 5/16 C. 5/8 D. 4/15

Answer 1

A is the one I think of

Answer 2

Answer:

A. 1/2 = 0.5

Step-by-step explanation:

We are given that the couple stops bearing a child until they get two boys.

Now, if the couple bears less than 2 girls.

Then the possible cases are: BB, BGB and GBB.

As the probability of the child being a boy = 0.5

Thus, the probability of the child being a girl = 1 - 0.5 = 0.5.

Hence, the probability that the couple bears less than 2 girls = sum of probabilities of BB, BGB and GBB.

i.e. Probability of less than 2 girls = $(\frac{1}{2})^{2}$ + $(\frac{1}{2})^{3}$ + $(\frac{1}{2})^{3}$

i.e. Probability = $\frac{1}{4} +\frac{1}{8} +\frac{1}{8}$

i.e. Probability = $\frac{1}{4} +\frac{2}{8}$

i.e. Probability = $\frac{1}{4} +\frac{1}{4}$

i.e. Probability = $\frac{2}{4}$

i.e. Probability = $\frac{1}{2}= 0.5$

Thus, the probability of less than 2 girls = 0.5

So, the probability of atleast 2 girls = 1 - Probability of less than 2 girls

i.e. The probability of atleast 2 girls = 1 - 0.5 = 0.5

Hence, the probability that the couple has atleast 2 girls is 0.5 = 1/2.